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QUESTION The number of cars passing along a road in 3 different
five minute periods are recorded as $n_1,n_2$ and $n_5$.  They may
be assumed to be independent observations from Poisson
distributions with means $\mu,\lambda\mu$ and $\lambda^2\mu$
respectively.  Show that the maximum likelihood estimates $\hat
\lambda$ and $\hat \mu$ satisfy:

\begin{eqnarray*} \hat \mu + \hat \lambda \hat \mu + (\hat
\lambda)^2\hat \mu & = & n_1+n-2+n_3\\ \hat \lambda\hat \mu+2(\hat
\lambda)^2\hat \mu & = & n_2+2n_3 \end{eqnarray*}

Find $\hat \lambda$ and $\hat \mu$ when $n_1=30,\ n_2=40$ and
$n_3=50.$




ANSWER
 $L(\lambda,\mu)=\frac{e^{-\mu}\mu^{n_1}}{n_1!}\frac{e^{-\lambda
  \mu}(\lambda \mu)^{n_2}}{n_2!}\frac{e^{-\lambda^2 \mu}(\lambda^2
  \mu)^{n_3}}{n_3!}\\
  \ln L(\lambda, \mu)=k-(\mu+\lambda \mu+\lambda^2
  \mu)+(n_1+n_2+n_3)\ln \mu +(n_2+2n_3)\ln \lambda\\
  \frac{\partial \ln L(\lambda, \mu)}{\partial \lambda}=-\mu
  -2\lambda \mu + \frac{n_2+2n_3}{\lambda}\\
  \frac{\partial \ln L(\lambda, \mu)}{\partial \mu}=-1-\lambda
  -\lambda^2 +\frac{n_1+n_2+n_3}{\mu}$\\
  hence mle's satisfy
  $\hat{\lambda}\hat{\mu}+2\hat{\lambda}^2\hat{\mu}=n_2+2n_3\\
  \hat{\mu}+\hat{\lambda}\hat{\mu}+\hat{\lambda}^2\hat{\mu}=n_1+n_2+n_3$


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